# Integrating by parts

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• November 9th 2009, 08:49 PM
C.C.
Integrating by parts
I have a problem that I've been working on:

Consider the definite integral http://hosted2.webwork.rochester.edu...201ea3f281.png

I have:

arcsin(4x)(x^2/2) minus the integral [IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot-1.jpg[/IMG][IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot-2.jpg[/IMG]http://hosted2.webwork.rochester.edu...eb778192f1.png

I'm really not sure where to go from here.
• November 9th 2009, 09:27 PM
redsoxfan325
Quote:

Originally Posted by C.C.
I have a problem that I've been working on:

Consider the definite integral http://hosted2.webwork.rochester.edu...201ea3f281.png

I have:

arcsin(4x)(x^2/2) minus the integral [IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot-1.jpg[/IMG][IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot-2.jpg[/IMG]http://hosted2.webwork.rochester.edu...eb778192f1.png

I'm really not sure where to go from here.

If $x=\frac{1}{4}\sin(u)$, then $dx=\frac{1}{4}\cos(u)$

Since $u=\arcsin(4x)$, the new bounds are $0...\frac{\pi}{2}$

So we have $\frac{1}{16}\int_0^{\pi/2} u\sin u\cos u\,du=\frac{1}{32}\int_0^{\pi/2}u\sin2u\,du$

From here it's an easy integration by parts. I calculate the final answer to be $\frac{\pi}{128}$.